2
$\begingroup$

I am stuck with showing that this function is differentiable at $x=0$ but not at $x=1$.

$f(x) = \begin{cases} x^2, & \text{if x}\in{\mathbb{Q}}\\ x^3, & \text{if x}\notin{\mathbb{Q}}\\ \end{cases}$

So for $x=0$:

Case 1: $h \in \mathbb{Q}$

$\lim\limits_{h \to 0}{\frac{f(0+h)-f(0)}{h}} = \lim\limits_{h \to 0}{\frac{f(h)}{h}} = \lim\limits_{h \to 0}{\frac{h^2}{h} = \lim\limits_{h \to 0}{ h} = 0} $

Case 2: $h \notin \mathbb{Q}$

$\lim\limits_{h \to 0}{\frac{f(0+h)-f(0)}{h}} = \lim\limits_{h \to 0}{\frac{f(h)}{h}} = \lim\limits_{h \to 0}{\frac{h^3}{h} = \lim\limits_{h \to 0}{ h^2} = 0} $

I am unsure on how to continue from here. I thought I need to show that for $x=0$ the right and left hand side limits are the same, and for $x=1$ they are different.

$\endgroup$
2
$\begingroup$

For $\;x=1\;$ we'd get

$$\frac{f(1+h)-f(1)}h=\begin{cases}&\frac{(1+h)^2-1}h=\frac{2h+h^2}h=(2+h)\xrightarrow[h\to0]{}2,&h\in\Bbb Q\\{}\\ &\frac{(1+h)^3-1}h=\frac{3h+3h^3+h^3}h=(3+3h+h^2)\xrightarrow[h\to0]{}3,&h\notin\Bbb Q\end{cases}$$

$\endgroup$
  • 2
    $\begingroup$ Oh I see... I simply forgot to do the same thing for $x=1$. Here we can see that they are not the same. Thanks! $\endgroup$ – ffritz Apr 20 '17 at 12:53
  • $\begingroup$ @Warhost Indeed...yet in zero the limits are equal and thus the function's differentiable there. No need to do one-sided limits in this case. $\endgroup$ – DonAntonio Apr 20 '17 at 12:54
1
$\begingroup$

Hint:

for $x=1$ the limit of the quotient goes to $2$ for rational $h$ and goes to $3$ for irrational $h$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.